Q. A 1m long steel wire of cross-sectional area 1 $mm^{2}$ is extended by 1 mm. If $Y = 2\ \times 10^{11}N\ m^{- 2}$ , then the work done is
A) 0.1 J
B) 0.2 J
C) 0.3 J
D) 0.4 J
Explanation
(a) $Y = \frac{Fl}{A\mathrm{\Delta}l}\ \ \ or\ \ F = \frac{YA\mathrm{\Delta}l}{l}$
Work done = $\frac{1}{2}\ F\mathrm{\Delta}l\ $
= $\frac{1}{2}\frac{FA(\mathrm{\Delta}l)^{2}}{l} = \frac{YA(l)^{2}}{2l}$
= $\frac{2 \times 10^{11} \times 10^{- 6} \times 10^{- 6}}{2\ \times 1} = 0.1\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.